More Probability Help?

Question

The problem is, well, I don't really know what pdf means.

The pdf of X is f(x)=kx^2 for 0 < x < 2 and f(x)=0 elsewhere.

Find the value of k.

I think what I have to do is find the value of k that yields a parabola that does... something. But I don't really know. Help?

Yeah, that's my problem, that's all the information I'm given.

To the idiot who said to talk to the teacher: maybe when you get to college you'll realize that in a class with 300 people, getting time with the teacher's a bit difficult.

Answer 1

You should integrate the pdf over the entire interval ranging from -inf to +inf and let the integral be 1,

___Integral{ f(x) dx; -inf ___= Integral{ 0 dx; -inf ___= 0+ (kx^3/3||2- kx^3/3||0)+ 0
___= 1

___k (8/3- 0) = 1
___k = 3/8
--- --- ---

The pdf is related to the probability of an event X between the values "a" and "b" to occur. This relationship is given by
___P(a
This is particularly useful when the events X's are said to be "continuous". Examples of "continuous" events include the lifespan of a lightbulb, the length of a pipe etc., which are usually constrained by some standard or quality issues.

The biggest difference between "continuous" and discrete probability is the amount of possible events. In a dice game, for instance, you have 6 different events: "1", "2"... "6". Lightbulbs, on the other hand, might last 1001, 1002, 1020, 1007 hours, which means that the number of events is much higher, and in this case is particularly defined according to the precision of the measuring equipment.

Answer 2

The probability density function is the equation relating each possible probaility for each value. In the probaility density function you have written above the value are between 0 and 2, not inlcuding 0 or 2. The probaility of any value being outside of 0 and 2 is zero. I think you would need more information to find out what value K is exactly. X^2 is a parabola and if K is negative than there is an inverted parabola but that is most I can know with the information you provided.

Answer 3

Integrate the pdf over the range of 2, 0 and set it equal to 1. Solve for k and you're golden.


Int kx^2 dx = 1/3 k * x^3 |{2, 0}] = 8/3 k = 1==> k = 3/8

Answer 4

hi...I understand your predicament, especially in a large class. However, when you (your parents) pay good money for tuition, you are entitled at least to some TA attention.

Enough of that...let me give you a hand on this one.

pdf = probability density function
range = 0
Recall that the integration of the probability density function over the range is equal to 1.

So the integral over the range is given by:

1 = integration [0...x...2] of kx^2 dx

= k/3*(x^3) evaluated between 0 and 2

1 = k/3 *(8-0)

solving for k = 3/8

Please follow these steps to convince yourself this is the correct result. Good luck!

Answer 5

No offense man but it's called a "TUTOR" otherwise known as a "TEACHER" you know that person your suppose to be listening to in class instead of (and we're all guilty of it) GOOFING off....but for real if your having trouble just explain to your teacher that your not catching on as quickly as you'd hoped and need to be refreshed on the assignment...after all I don't know anyone who actually studies over summer vacation do you??

Answer 6

read this
http://en.wikipedia.org/wiki/Probability_density_function

pdf = probability density function


And you will need more information to get k. (don't listen to the previous answer, he doesn't know what he's talking about)